How many n-letter words are there, such that number of letters “a” is even? [closed]












-1












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How many n-letter words (made of letters from 25-letter english alphabet) are there, such that number of letters "a" is even? ("a" appears even number of times in a word).




I'm trying to create recursive formula, but with no success.










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closed as off-topic by Did, mrtaurho, Saad, user91500, Lord_Farin Dec 24 '18 at 9:07


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, mrtaurho, Saad, user91500, Lord_Farin

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    If we have a "good" word of length $n-1$, how many letters can we append to make a good word of length $n$? Same question if we have a "bad" word of length $n-1$.
    $endgroup$
    – lulu
    Dec 20 '18 at 13:08










  • $begingroup$
    What happened to the missing letter?
    $endgroup$
    – paw88789
    Dec 20 '18 at 15:56










  • $begingroup$
    The same question, with two other letters instead of 24, is here
    $endgroup$
    – Ross Millikan
    Dec 20 '18 at 16:06










  • $begingroup$
    This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
    $endgroup$
    – Did
    Dec 23 '18 at 21:34
















-1












$begingroup$



How many n-letter words (made of letters from 25-letter english alphabet) are there, such that number of letters "a" is even? ("a" appears even number of times in a word).




I'm trying to create recursive formula, but with no success.










share|cite|improve this question











$endgroup$



closed as off-topic by Did, mrtaurho, Saad, user91500, Lord_Farin Dec 24 '18 at 9:07


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, mrtaurho, Saad, user91500, Lord_Farin

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    If we have a "good" word of length $n-1$, how many letters can we append to make a good word of length $n$? Same question if we have a "bad" word of length $n-1$.
    $endgroup$
    – lulu
    Dec 20 '18 at 13:08










  • $begingroup$
    What happened to the missing letter?
    $endgroup$
    – paw88789
    Dec 20 '18 at 15:56










  • $begingroup$
    The same question, with two other letters instead of 24, is here
    $endgroup$
    – Ross Millikan
    Dec 20 '18 at 16:06










  • $begingroup$
    This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
    $endgroup$
    – Did
    Dec 23 '18 at 21:34














-1












-1








-1





$begingroup$



How many n-letter words (made of letters from 25-letter english alphabet) are there, such that number of letters "a" is even? ("a" appears even number of times in a word).




I'm trying to create recursive formula, but with no success.










share|cite|improve this question











$endgroup$





How many n-letter words (made of letters from 25-letter english alphabet) are there, such that number of letters "a" is even? ("a" appears even number of times in a word).




I'm trying to create recursive formula, but with no success.







combinatorics recursion combinatorics-on-words






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share|cite|improve this question













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share|cite|improve this question








edited Dec 20 '18 at 15:34









greedoid

42.9k1153105




42.9k1153105










asked Dec 20 '18 at 13:05









lellerleller

715




715




closed as off-topic by Did, mrtaurho, Saad, user91500, Lord_Farin Dec 24 '18 at 9:07


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, mrtaurho, Saad, user91500, Lord_Farin

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Did, mrtaurho, Saad, user91500, Lord_Farin Dec 24 '18 at 9:07


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Did, mrtaurho, Saad, user91500, Lord_Farin

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    If we have a "good" word of length $n-1$, how many letters can we append to make a good word of length $n$? Same question if we have a "bad" word of length $n-1$.
    $endgroup$
    – lulu
    Dec 20 '18 at 13:08










  • $begingroup$
    What happened to the missing letter?
    $endgroup$
    – paw88789
    Dec 20 '18 at 15:56










  • $begingroup$
    The same question, with two other letters instead of 24, is here
    $endgroup$
    – Ross Millikan
    Dec 20 '18 at 16:06










  • $begingroup$
    This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
    $endgroup$
    – Did
    Dec 23 '18 at 21:34


















  • $begingroup$
    If we have a "good" word of length $n-1$, how many letters can we append to make a good word of length $n$? Same question if we have a "bad" word of length $n-1$.
    $endgroup$
    – lulu
    Dec 20 '18 at 13:08










  • $begingroup$
    What happened to the missing letter?
    $endgroup$
    – paw88789
    Dec 20 '18 at 15:56










  • $begingroup$
    The same question, with two other letters instead of 24, is here
    $endgroup$
    – Ross Millikan
    Dec 20 '18 at 16:06










  • $begingroup$
    This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
    $endgroup$
    – Did
    Dec 23 '18 at 21:34
















$begingroup$
If we have a "good" word of length $n-1$, how many letters can we append to make a good word of length $n$? Same question if we have a "bad" word of length $n-1$.
$endgroup$
– lulu
Dec 20 '18 at 13:08




$begingroup$
If we have a "good" word of length $n-1$, how many letters can we append to make a good word of length $n$? Same question if we have a "bad" word of length $n-1$.
$endgroup$
– lulu
Dec 20 '18 at 13:08












$begingroup$
What happened to the missing letter?
$endgroup$
– paw88789
Dec 20 '18 at 15:56




$begingroup$
What happened to the missing letter?
$endgroup$
– paw88789
Dec 20 '18 at 15:56












$begingroup$
The same question, with two other letters instead of 24, is here
$endgroup$
– Ross Millikan
Dec 20 '18 at 16:06




$begingroup$
The same question, with two other letters instead of 24, is here
$endgroup$
– Ross Millikan
Dec 20 '18 at 16:06












$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Did
Dec 23 '18 at 21:34




$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Did
Dec 23 '18 at 21:34










1 Answer
1






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1












$begingroup$

Let $o_n$ be a number of n-lenght words with odd number of $a$ and



let $e_n$ be a number of n-lenght words with even number of $a$.



Then $o_n+e_n = 25^n$ and $$ e_{n+1} = o_n+ 24e_n$$



that is if first letter is $a$ then in the rest of a word must be odd number of $a$ and if first letter is not $a$ then the number of even $a$ is the same as in a $n$-lenght word times 24 (since we have 24 choises for first number) .






share|cite|improve this answer











$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Let $o_n$ be a number of n-lenght words with odd number of $a$ and



    let $e_n$ be a number of n-lenght words with even number of $a$.



    Then $o_n+e_n = 25^n$ and $$ e_{n+1} = o_n+ 24e_n$$



    that is if first letter is $a$ then in the rest of a word must be odd number of $a$ and if first letter is not $a$ then the number of even $a$ is the same as in a $n$-lenght word times 24 (since we have 24 choises for first number) .






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Let $o_n$ be a number of n-lenght words with odd number of $a$ and



      let $e_n$ be a number of n-lenght words with even number of $a$.



      Then $o_n+e_n = 25^n$ and $$ e_{n+1} = o_n+ 24e_n$$



      that is if first letter is $a$ then in the rest of a word must be odd number of $a$ and if first letter is not $a$ then the number of even $a$ is the same as in a $n$-lenght word times 24 (since we have 24 choises for first number) .






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $o_n$ be a number of n-lenght words with odd number of $a$ and



        let $e_n$ be a number of n-lenght words with even number of $a$.



        Then $o_n+e_n = 25^n$ and $$ e_{n+1} = o_n+ 24e_n$$



        that is if first letter is $a$ then in the rest of a word must be odd number of $a$ and if first letter is not $a$ then the number of even $a$ is the same as in a $n$-lenght word times 24 (since we have 24 choises for first number) .






        share|cite|improve this answer











        $endgroup$



        Let $o_n$ be a number of n-lenght words with odd number of $a$ and



        let $e_n$ be a number of n-lenght words with even number of $a$.



        Then $o_n+e_n = 25^n$ and $$ e_{n+1} = o_n+ 24e_n$$



        that is if first letter is $a$ then in the rest of a word must be odd number of $a$ and if first letter is not $a$ then the number of even $a$ is the same as in a $n$-lenght word times 24 (since we have 24 choises for first number) .







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 20 '18 at 13:30

























        answered Dec 20 '18 at 13:08









        greedoidgreedoid

        42.9k1153105




        42.9k1153105















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